Transcript Chapter 15

Immediate Consequences of Symmetry
I. Optical Activity (Chirality)
• If a mirror image of a molecule cannot be
superimposed on the original the molecule is chiral.
• Chiral molecules are optically active in the sense
that they rotate the plane of polarized light.
• For complex molecules this is difficult to visualize
and symmetry elements may assist in the
determination of optical activity.
Cannot Have Sn
• A molecule which contains an n-fold improper
axis is always superimposable on its mirror
image. Recall that Sn consists of a rotation
followed by a reflection.
• Since a reflection creates the mirror image, Sn is
equivalent to rotating in space the mirror image.
By definition, a molecule containing an Sn axis is
brought into coincidence with itslef by the Sn
operation and hence its mirror image, after
rotation, is superimposable.
Watch out!
• It follows from symmetry that a molecule may
only be chiral if it does not possess an improper
rotation axis, Sn .
• This element may be implied by other elements
present. Recall that S2 is equivalent to i . Also,
S1 is equivalent to s.
• Therefore, molecules with any one of s, i, or Sn
cannot be optically active.
2. Dipole moment
• A polar molecule is one with a permanent
electric dipole moment.
• This can be the case only if the center of
negative and positive charges do not coincide.
• Since a symmetry operation leaves a molecule
in a configuration physically indistinguishable
from before, the direction of the dipole moment
vector must also remain unchanged.
Must coincide with symmetry
elements
• The direction of the dipole moment must be
coincident with each of the symmetry elements.
It cannot be perpendicular to any mirror plane or
axis of rotation.
• If a molecule has a Cn axis, the dipole must lie
along this axis. If there is a s, it must lie in this
plane; if several planes, it must lie at their
intersection. E.g., for NH3, dipole moment is
along C3 axis, which is also intersection of 3 sv.
Limits point groups with dipoles
•
It follows that molecules with an i (operation
reverses direction of vectors) or with two or more
non-coincident Cn axes (dipole moment cannot lie
on two axes at the same time) cannot have a dipole
moment (this rules out the D and higher symmetry
families).
• Only molecules belonging to Cn , Cnv , or Cs may
have a dipole moment. For Cn and Cnv , the dipole
moment must lie along the symmetry axis, and n
may be any value from 1 to .
• Thus, we must include C1 and Cv in these
categories.
In-class activity
• Which of the following are:
a) polar
b) optically active
• CCl4
CHCl3
• H2O2 (skew)
CO
CO2
H2O2 (staggered)
Point Groups and Multiplication
Tables
• A point group is a set of symmetry operations that
form a complete multiplication table containing all
products and reciprocals of its elements.
• Multiplication of two symmetry operations is
defined as a sequence of symmetry operations.
– The product C4s means “do the s operation and
then do the C4 on the figure that results from
doing the s operation.”
– Note that the order of execution of the
symmetry operations is from right to left.
• Every multiplication table must contain the
identity.
• Every symmetry operation must have a
reciprocal which is defined as the symmetry
operation done in reverse.
• The product of a symmetry operation and its
reciprocal is the identity operation.
Definition of a group
•
A group is any collection of elements which
together with some well-defined combining
operation (ordinary algebra, matrix algebra, one
operation followed by another, etc.) obey a
certain set of rules.
1. The ‘product’ or combination of any two
elements in a group must produce an element
which is also in the group.
2. The group must contain the identity element,
which combines with any element in the group,
leaving that element unchanged.
3. The associative law must hold for all elements of
the group.
4. Every element must have an inverse (combined,
they yield identity) which is also a member of the
group.
1. PQ = R
R in the group
2. RE = ER = R
R in the group
3. P(QR) = (PQ)R
For all elements
4. RR-1 = R-1R = E
R-1 in the group
Order and Classes
• The number of symmetry operations in a group is
called the order of the group, h.
• A group always consists of at least two
subgroups, each of which satisfies the
requirements for a group.
– The identity is always a subgroup of order 1.
– The order, g, of a subgroup, is always an
integral divisor of the order, h, of the group to
which it belongs.
• A group always consists of two or more classes,
each of which contains symmetry operations
which can be transformed into one another by a
symmetry operation of the group.
Note that the
separate
existence of a
C3 and a sh
requires the
existence of
an S3 axis.
C3, S3
C2, sv
The C3 and C32 operations
are in the same class.
F
B
F
The S3 and S35 operations
are in the same class.
C2, sv
F
C2, sv
Molecular plane is sh.
The three C2 axes are equivalent so the C2 operations are in the same
class.
The three sv planes are equivalent so the sv operations are in the same
class.
Rules For Symmetry Operations and
Classes
• The Inversion:
– There is only one inversion and it is always in a
class by itself.
• Reflections:
– Reflection in a plane perpendicular to the axis of
highest symmetry is designated sh and forms a
separate class.
– Reflection in a plane that contains the axis of
highest symmetry and passes through several
atoms is designated sv. All are in the same class.
– Reflection in a plane that contains the axis of
highest symmetry and passes between sets of
atoms is designated sd. All are in the same class.
Proper Rotations:
– In the Cn point groups, Cn, Cn2, Cn3, etc
are in separate classes.
– In groups of higher symmetry, Cnm and
Cnn-m are in the same class.
Systematic Generation of Point
Groups
• We will examine the generation of some of the
simpler point groups by constructing multiplication
tables for symmetry operations.
• Each point group is generated by starting with the
identity and at least one other symmetry operation.
• The additional symmetry operations needed to
complete the point group will arise automatically
since all products must be members of the group.
• Multiplication is not necessarily commutative (AB is
not necessarily the same as BA).
• Multiplication is always associative (A[BC] = [AB]C).
• No two rows and no two columns in a multiplication
table can be the same.
Groups of Low Symmetry
No symmetry: Only the identity is a symmetry operation.
Schönflies symbol
for the point group.
C1
E
E
E
The designation of this as the C1 point
group means that the identity operation is
considered as a C1 rotation axis. Since C1
is rotation by 2/1, this corresponds to
doing nothing to the object.
sh
Cs
E
E
E
sh
sh
E
sh
Note that, like the number 1,
the identity multiplied by
anything else leaves the thing it
is multiplied by unchanged.
Multiplication by the identity is
always commutative.
sh is reflection in the molecular plane.
Since there is only one reflection plane,
it is automatically the molecular plane.
E C3 C32
E E C3 C32
C3 C3 C32 E
C3
C32 C32 E
C3
Note that the product
C3C3 has generated
the new product C32
which must also be a
member of the group.
C32 C32 = four C3 operations in sequence
= C34 = C33C3 (associative law)
Since C33 = E, C32C32 = EC3 = C3
(multiplication by the identity doesn’t
change anything)
Vectors, Matrices and Point
Groups
• In the quantum description of a chemical bond, the
wavefunctions for the electrons in the bond have
the same directional properties as the bond.
• Since wavefunctions for the electrons in a bond
have directional properties, they can be treated as
vectors.
• Group Theory provides a systematic classification
of vectors in terms of their geometric properties.
• Any vector can be written as the product of a
column matrix of unit vectors and a square matrix
describing the geometric properties of the vector.
• The properties of this square geometric matrix
provide the quantitative link between Group Theory
and chemical bonding.
Representations of Point Groups
• Each symmetry operation has geometric properties that can
be expressed in a geometric transformation matrix that is
specific for that operation.
• A complete set of geometric transformation matrices for the
operations of a point group as applied to a particular object
is called the representation of the group.
• In many cases, the full geometric transformation matrix
consists of smaller square matrices arranged along the
diagonal of the larger matrix. Corresponding sets of these
smaller matrices are also representations of the group.
• Representations consisting of matrices that do not factor
into smaller ones are called irreducible representations.
• The dimension of a representation is the number of rows or
columns in its geometric transformation matrices.
a
b
c
0
0
0
0
0
0
d
e
f
0
0
0
0
0
0
g
h
i
0
0
0
0
0
0
0
0
0 j
k
0
0
0
0
0
0
0 l m
0
0
0
0
0
0
0 0
0 n
o
p
q
0
0
0 0
0 r
s
t
u
0 0
0 0
0 v
w
x
y
0
0 0
0 z



0
The Characters of Matrices
• Using the full geometric matrices is very cumbersome
– Each atom in a molecule requires three unit vectors
(three coordinates) to locate it.
– For a molecule containing n atoms the geometric matrix
would be 3n x 3n for each symmetry operation in the
group.
• It is much more convenient to describe the molecule in
terms of the characters of the geometric transformation
matrices.
• The character, , of a square matrix is the sum of its
diagonal elements.
• The representations of molecules are nearly always
written as the characters of the symmetry operations,
not as the actual geometric transformation matrices.
Notes
• The principal rotation axis is chosen as the
z-axis.
• There is always one completely symmetric
irreducible representation (each element is
1).
The Geometric Transformation Matrices
Reflection in a Plane:
The x and y unit vectors
are in the reflection plane
and are unchanged.
Initial Vector
x, y plane
The z unit vector is
perpendicular to the
plane and reverses.
Reflected Vector
x 2 = x1 + 0 y 1 + 0 z 1
x2
y2 = 0 x1 + y1 + 0 z1
y2
z2 = 0 x1 + 0 y1 – z1
z2
=
In general, for a reflection transform, all
off-diagonal elements are 0, on-diagonal
are 1 for coordinates in plane, -1 for the
other.
1 0 0
x1
0 1 0
y1
0 0 -1
z1
Transformation matrix
Inversion:
x2 = -x1 + 0 y1 + 0 z1
x2
y2 = 0 x1 - y1 + 0 z1
y2
z2 = 0 x1 + 0 y1 - z1
z2
=
-1 0 0
x1
0 -1 0
y1
0 0 -1
z1
0 if off-diagonal, -1 on-diagonal (all coordinates are
inverted).
Proper Rotation by an angle :
x2
y2
z2
=
cos  -sin  0
x1
sin 
y1
0
cos  0
0
1
z1
We now have non-zero off-diagonal
elements, which sometimes
complicates matters, which we’ll try to
avoid in this course.
Improper Rotation:
x2 = x1 cos  - y1 sin  + 0 z1
x2
cos  -sin  0
y2 = x1 sin  + y1 cos  + 0 z1
y2 = sin 
z2 = 0 x1 + 0 y1 – z1
z2
0
cos  0
0
-1
The only difference from the proper rotation is the effect on
the z vector of the reflection perpendicular to the rotation axis.
x1
y1
z1
Rules for Manipulating
Representations
1.
The sum of the squares of the dimensions (li2)
of the irreducible representations of a group is
equal to the order (h) of the group.
2
2
2
l

l

l
 i 1 2   h
all representations
2. The sum of the squares of the characters ([i(R)]2)
in any one irreducible representation of a group
equals the order (h) of the group. (R is any
symmetry operation,  is the character, i and j for
different representations)
 R
2
i
all operationsR
h
3. The vectors whose components are the characters [i(R)]
of two different irreducible representations (i and j) are
orthogonal.
  R R  0
i
all operations R
j
4. In any given representation, either reducible or
irreducible, the characters of all matrices belonging
to the same class are identical.
5. The number of irreducible representations in a
group is equal to the number of classes in the group.
Application to the C2v Point Group
• This point group has four symmetry operations,
E, C2, sv(x,z) and sv(y,z). Each is in a separate
class.
Number of irreducible representations = ?? (Rule 5) 4
Possible dimensions of representations are ?? (Rule 1) 1
• Since the order of the group is 4, the sum of the li2 = 4.
• There are four irreducible representations and the
smallest possible matrix is 1x1 so the smallest possible
value of l is 1.
• Only l = 1 will sum to 4 when squared and added
four times.
x2
1 0 0
x1
E y2
= 0 1 0
y1
z2
0 0 1
z1
For C2 :
Therefore
and
C2
θ = 180°
sin 180° = 0
cos 180° = -1
0
x2
y2
=
z2
x1
cos  0
0
y1
1
z1
-1 0 0
x1
0 -1 0
y1
0 0 1
z1
-1 0 0
x1
0 1 0
y1
0 0 1
z1
σv’(yz)
x2
z2
y2 = sin 
z2
σv(xz)
y2
cos  -sin  0
x2
=
1 0 0
x1
x2
0 -1 0
y1
y2
0 0 1
z1
z2
=
We now “block diagonalize” each transformation
matrix- break it down into smaller matrices including
the non-zero elements along the diagonal.
[1] 0 0
E
0 [1] 0
0
[-1] 0 0
C2
0 [1]
σv(xz)
0 [-1] 0
0
σv’(yz)
0 [1]
[1] 0 0
[-1] 0 0
0 [-1] 0
0
[1] 0
0
0
0 [1]
0 [1]
In this case the
x, y, z axes are
also block
diagonalized,
and can be
treated
independently.
Γ
sv(xz)
sv(yz)
coordinate
E
C2
1
-1
1
-1
x
1
-1
-1
1
y
1
1
1
1
z
3
-1
1
1
Here we have composed representations by simply
transcribing the element (the character of a 1x1 matrix)
in the 1,1 position of each transformation matrix, which
describes the result of the symmetry operation on the x
axis; and similarly for 2,2 (y) and 3,3 (z). By adding the
values in each column above, we can get a reducible
representation.
4th representation
•We obtain this by following all the rules:
- The character for the identity must be 1
- The sum of squares of the characters in
each representation must equal the order (4)
- No two representations may be the same
- The representations must all be
orthogonal to each other.
C2v
E
C2
sv(xz)
sv(yz)
1
1
1
1
1
2
1
-1
1
-1
3
1
1
-1
-1
4
1
-1
-1
1
The Structure of Character Tables
• The left-hand column gives the Schönflies symbol
for the point group with the Mulliken symbols for
the irreducible representations below it.
• The main part of the table has the symmetry
operations at the top, arranged in classes, with the
identity at the left followed by the other symmetry
operations starting with the highest order rotation
axis and followed in order of decreasing
symmetry. Below this are listed the characters for
the symmetry operations in each irreducible
representation.
• At the far right are two columns showing the
transformation properties of coordinates and their
squares and cross-products.
1.
2.
3.
4.
Rules for Assigning Mulliken
Symbols
All one-dimensional representations are either A
or B. Two-dimensional representations are E and
three-dimensional ones are T or F.
One-dimensional representations that are
symmetric (positive character) with respect to
rotation about the principal axis are A while those
that are antisymmetric (negative character) are B.
The subscript 1 designates an A or B
representation that is symmetric with respect to
rotation about a C2  Cn or, in its absence, is
symmetric with respect to a sv. The subscript 2
indicates antisymmetric behaviour.
A ‘ designates symmetry with respect to a sh. A “
indicates antisymmetric behaviour.
5. The subscript g (for gerade meaning
even in German) indicates symmetry
with respect to inversion. The subscript
u indicates antisymmetric behaviour.
The column containing x, y, z, Rx, Ry and Rz gives the
irreducible representations for translation along the
three Cartesian axes (x, y, z) and for rotation about
the three Cartesian axes (Rx, Ry, Rz).
The column at the far right containing the squares
and cross-products gives the irreducible
representations for properties (such as d-orbital
wavefunctions and molecular polarizability) that
depend on these squares and cross-products.
C2v
E
C2
A1
A2
B1
B2
1
1
1
1
1
1
-1
-1
sv(xz) sv(yz)
1
-1
1
-1
1
-1
-1
1
z
Rz
x, Ry
y, Rx
x2, y2, z2
xy
xz
yz
Groups with Classes Larger than 1
• Recall that the characters of all the matrices in a
class are identical. This means that
– Character tables can be compressed by lumping all
the operations in a class in a single column.
– The sums over symmetry operations required in
analyzing representations can be taken over
classes provided we multiply the product for a
given class by the order, g, of the class.
Rule 2:
Rule 3:
2





R
 i k gk  h
all classes, k
  R R g
i
all classes, k
j
k
k
0
Remember
that the class
order for each
symmetry
operation can
be different.
Consider the C3v Point Group
C3v
E
2C3
3sv
1
1
1
1
2
1
1
-1
3
2
-1
0
Rule 5: There are 3 classes so the number of irreducible
representations = ? 3
Rule 1: There are 6 operations so l2 = 6  l =?? 1, 1, 2
2





R
 i k gk  h
all classes, k
For the one dimensional representations,  can be?? ± 1
For the two dimensional representation, (E) has
to be?? 2 since each diagonal element has to be 1 (the
unit vectors are unchanged). This means that (C3) =??
-1 and (sv) =?? 0.
x2
1 0 0
E y2
= 0 1 0
y1
z2
0 0 1
z1
For C3 :
Therefore
and
2C3
θ = 120° (2/3)
sin 120° = √3/2
cos 120° = -1/2
3σv(xz)
x2
y2
z2
=
cos  -sin  0
x2
x1
1 0 0
x1
0 -1 0
y1
0 0 1
z1
y2 = sin 
x1
cos  0
z2
0
x2
-1/2 -√3/2 0
x1
-1/2 0
y1
y2
z2
=
√3/2
0
0
y1
0
1
1
z1
z1
The transformation matrix for C3 cannot be block
diagonalized into 1x1 because it has off-diagonal
elements. It can be blocked into 2x2 and 1x1
matrices; the others must follow the same pattern for
consistency.
-1/2 -√3/2 0
1 0 0
E
0 1 0
0
0 [1]
3σv(xz)
1
0 0
0 -1 0
0
0 [1]
2C3
√3/2
0
-1/2 0
0
[1]
The C3 matrix must be
blocked this way because
the (x,y) combination is
needed for the new x,y
coordinates.
Building the representations
C3v
E
2C3
3sv
1
1
1
1
2
1
1
-1
3
2
-1
0
coordinate
z
x, y
Here we will compose the 2-D representation by
taking the characters of the 2x2 matrices (summing
along the diagonal), and one of the 1-D
representations from the characters of the 1x1
matrices.
The third representation can be found by
using the defining properties of group tables.
Practice problem
• For the D2d point group (character table on next
slide):
• A) determine the order of the group
• B) test the E representation for orthogonality
with the other 4 representations (Rule 3)
• C) Confirm that the sum of the squares of the
characters in each irreducible representation
equals the order of the group (Rule 2)
D2d
A1
A2
B1
B2
E
E
1
1
1
1
2
2S4
1
1
-1
-1
0
C2
2C2’
1
1
1 -1
1
1
1 -1
-2
0
2sd
1
-1
-1
1
0
Reducing Reducible
Representations
• Virtually every practical application of group
theory starts with a reducible representation.
• All reducible representations contain some
integer multiple of each of the irreducible
representations of the point group.
• The application of group theory to problems
begins by determining how many of each kind of
irreducible representation are contained in the
reducible representation.
• The irreducible representations can be thought of
as being like components of a vector and
determine the geometric features of the
phenomenon being analyzed.
• We could construct a reducible representation for a
point group by multiplying each irreducible
representation by some integer, a, and adding the result
for each symmetry operation, R, over all irreducible
representations. The character of each symmetry
operation in the reducible representation would then be
 R    ai  i R 
irreducible
representations
• We normally start with a knowledge of (R) in the
reducible representation and need to determine the value
of ai for each irreducible representation in the group.
• Since irreducible representations are orthogonal, if we
multiply the characters in the reducible representation by
those in a given irreducible representation, only the terms
for the “test” irreducible representation will survive.
These observations can be summarized in the equation
1
ai    R i R g R
h R
in which (R) is the character for symmetry
operation R in the reducible representation and i(R)
is the character for the same symmetry operation in
the ith irreducible representation. ai is the number of
times the ith irreducible representation is repeated in
the reducible one.
A C3v Example
C3v
E
2C3
3sv
1
1
1
1
2
1
1
-1
3
2
-1
0
1
-3
reducible
7
1
 E 1 E   2 C3 1 C3   3 s v 1 s v 
6
1
 7 1  211  3- 3 1
6
7  2 - 9  0

6
For 1: a1 
For 2:
1
a2  7 1  21 1  3- 3  -1
6
18
 3
6
For 3:
a3 
1
7  2  21 -1  3- 3  0
6
12

2
6
Application to Molecular Vibrations
and Infrared and Raman
Spectroscopy
• For a molecule containing n atoms, 3n
coordinates (component vectors) are needed to
completely define the positions of all the atoms.
• Translational motion of the molecule through
space accounts for three coordinates,
representing the location of the center of mass.
• Rotational motion of the molecule accounts for
– Two coordinates if the molecule is linear.
– Three coordinates if the molecule is not linear.
• The remaining coordinates, (3n-5) or (3n-6), are
required to describe vibrations of the atoms
relative to each other.
Probing Structure With Infrared
Spectroscopy
• When a molecule vibrates, its dipole
moment may oscillate.
• This oscillating dipole moment interacts with the electric
vector of electromagnetic radiation (light).
• Only those vibrations which have the same irreducible
representations as translational motion along x, y, or z will
interact with light. These will be vibrations which cause an
oscillation in the (temporary) dipole moment.
• The reducible representation of the molecule contains the
relation between molecular structure and this interaction.
• When a molecular vibration has the same frequency
as the light shining on it, the vibration will absorb energy
from the light beam.
• The frequency (energy) at which the light is absorbed
depends on the strength of the vibrating bond and the
masses of the atoms involved in the bond.
• The energies of molecular vibrations are of the same
magnitude as the energy of light in the infrared part of the
spectrum.
• The number of peaks in the infrared spectrum of a
molecule can be determined by using group theory and
this provides a link between molecular structure and the
infrared spectrum.
The Recipe
• Determine the reducible representation for all motions
of the molecule.
– Attach three unit vectors (x, y, z) to each atom.
– For each symmetry operation, only those atoms
that do not change position contribute to the
character for that symmetry operation in the
reducible representation.
– A unit vector that does not change its orientation
contributes +1 to the character in the reducible
representation.
– A unit vector that reverses its direction contributes
–1.
•For Cn operations, the sum of the diagonal
elements of the geometric matrix is always
1+2cos per unshifted atom for rotation by
the angle .
•For Sn operations, the sum of the diagonal
elements of the geometric matrix is always
–1+2cos  per unshifted atom for rotation by
the angle .
• Reduce the reducible representation to find the
number of each kind of irreducible representation
it contains.
• Subtract the irreducible representations for
translation (x, y, z) and rotation (Rx, Ry, Rz)
leaving those for the vibrations.
The Ammonia Molecule
The ammonia molecule belongs to the C3v point group.
z
z
sv
x
H
x
y
y
N
z
z
x
H
x
H
sv
y
y
sv
C3
E:- There are 12 vectors all of which are unchanged
by E so (E) = 12.
C3:- Only N is unshifted by C3. The rotation angle is 2/3
so the character for C3 is 1+2cos(2/3) = 1 + 2x(-0.5) = 0.
sv:- For any reflection plane, the N and one H are unshifted. Since the character per atom for sv is 1 (z and x
stay the same, y changes sign), the total character for the
two atoms is 2.
The reducible representation is
C3v
E
2C3
3sv
A1
A2
E
1
1
2
1
1
-1
1
-1
0
12
0
2
red.
This factors into the following irreducible representations.
1
18
n( A1 )  12 1  2  0 1  3  2 1 
3
6
6
1
6
n A2   12 1  2  0 1  3  2  -1   1
6
6
1
24
nE   12  2  2  0  -1  3  2  0  
4
6
6
red = 3A1 + 1A2 + 4E
red = 3A1 + 1A2 + 4E
Translation contributes A1 + E (remember that E is doubly
degenerate so one E accommodates translation along x and y).
Rotation contributes A2 + E (again E is doubly degenerate
and so includes both Rx and Ry).
The irreducible representations for the vibrations of NH3
are therefore 2A1 + 2E.
The
of vibrational
C3vnumber
E
2C
3sv modes is 2 (from A1) plus 2x2 = 4
3
(from the doubly degenerate E) for a total of 6. Since the
2 , z2
z
1
1 number
x2 + yis
A1 1 is nonlinear,
molecule
the
of vibrations
3x4 – 6 = 6
1
1 the-1
A2 agrees
which
with
groupRtheory
calculation as required.
z
2 - y2, xy) (xz, yz)
(x,y)
(R
,
R
)
(x
2
-1
0
E
x
y
Infrared active transitions have to transform like x, y, or z so
we conclude that all these vibrations are infrared active.
The A1 vibrations are symmetric with respect to both C3
and sv and so describe the symmetric N-H bend and the
symmetric N-H stretch.
N
H
H
N
H
H
Symmetric Stretch
H
H
Symmetric Bend
The two doubly-degenerate E vibrations are
antisymmetric with respect to the C3 axis. They
describe asymmetric bends and stretches.
N
H
H
N
H
H
Asymmetric Stretch
H
H
Asymmetric Bend
Practice Problem
• Using the x, y, and z coordinates for each
atom in CH4, determine the reducible
representation; reduce it; classify the
irreducible representations into
translational, rotational and vibrational
modes; and decide which vibrational
modes are IR active.
Analysis of Specific Vibrations
Suppose we want to find the number of infrared active
stretching vibrations in Ni(CO)4. The molecule is
tetrahedral as shown below.
We start by drawing vectors pointing in
the direction of the stretch along each
bond axis.
CO
Ni
CO
CO
CO
Td Point
Group
The vibrations will have the
transformation properties of these
vectors so we deduce the reducible
representation for this set of vectors in
the normal way and then decompose
it to find what irreducible
representations it contains.
CO
C2,S4,sd
Ni
CO
C3
CO
CO
Td
E 8C3
3C2
6S4
6sd
A1
1
1
1
1
1
A2
1
1
1
-1
-1
E
2
-1
2
0
0
T1
3
0
-1
1
-1
(Rx,Ry,Rz)
T2
3
0
-1
-1
1
(x, y, z)
vib 4
1
0
0
x2 + y2 + z2
(2z2-x2-y2, x2-y2)
(xy, xz, yz)
2
E leaves all four vectors in place and unchanged in direction so it
contributes +4.
C3 leaves one vector in place and unchanged in direction so it
contributes +1.
C2 and S4 move all the vectors and so do not contribute.
sd leaves two vectors in place and unchanged in direction so it
contributes +2.
Next decompose the reducible representation.
Td
E 8C3
3C2
6S4
6sd
A1
1
1
1
1
1
A2
1
1
1
-1
-1
E
2
-1
2
0
0
T1
3
0
-1
1
-1
(Rx,Ry,Rz)
T2
3
0
-1
-1
1
(x, y, z)
vib 4
1
0
0
x 2 + y2 + z2
(2z2-x2-y2, x2-y2)
(xy, xz, yz)
2
1
4 1  8 11  0  0  6  2 1  24  1
24
24
1
1
n A2  4 1  8 11  6  2  (-1)  0 nT1  4  3  8 1 0  6  2  (-1)  0
24
24
1
1
nE  4  2  8 1 (-1)  6  2  0  0 nT2  4  3  8 1 0  6  2 1  1
24
24
n A1 
There are 1 A1 and 1 triply degenerate T2 (four vibrations).
Only T2 transforms like x, y, z so only it is infrared active.
Raman Spectroscopy
• Raman effect depends on change in polarisability .
(See far-right column in character tables; explicitly
given in Hollas spectroscopy text).
• General Rule: If a vibrational mode has the same
symmetry as one or more of the binary combinations
of x, y and z the transition from this mode will be
Raman active.
• Infrared and Raman are based on two DIFFERENT
phenomena and therefore there is no necessary
relationship between the two activities.
• The higher the molecular symmetry the fewer “coincidences” between Raman and infrared active
modes.
x2, y2, z2, xy, xz and yz
For C2v
Transform as:
A1, A1, A1, A2, B1 and B2
C2v
E
C2
A1
A2
B1
B2
1
1
1
1
1
1
-1
-1
sv(xz) sv(yz)
1
-1
1
-1
1
-1
-1
1
z
Rz
x, Ry
y, Rx
x2, y2, z2
xy
xz
yz
Therefore, any vibrational mode in a C2v molecule
is Raman active (while A1, B1, and B2 modes are also IR
active).
2C3
3sv
C3v
E
A1
1 1
1
Tz
A2
1 1
-1
Rz
E
2 -1
0
(Tx,Ty) or
(Rx,Ry)
x2 + y2, z2
(x2 - y2, xy)
(yz, xz)
For C3v, only A1 and E modes would be
Raman active (same for IR activity).
Td
E
8C3
3C2
6S4
6sd
A1
1
1
1
1
1
A2
1
1
1
-1
-1
E
2
-1
2
0
0
T1
3
0
-1
1
-1
(Rx,Ry,Rz)
T2
3
0
-1
-1
1
(x, y, z)
x2 + y2 + z2
(2z2-x2-y2, x2-y2)
(xy, xz, yz)
Self-test: Which modes in a tetrahedral
molecule are Raman active? Which are IR
active?